Do we have that $a_{ij}\in \{0,1\}$?

how is $a_1=2$?

Please answer the question of Arthur. If the entries of your matrices are for example any integer, there is an infinite number of such matrices...

@ Phicar, $a_1=1$.

Suppose $a_{ij} = 1$. By the first condition, $a_{ij} = 1$ implies $a_{ji} = 0$. But if $a_{ji} = 0$, then $a_{ij} = 0$ due to the second condition. Is there something wrong with the conditions?

Excuse my typos, I think I have retyped the things correctly. APologies. Kindly have another look.

This seems similar to your previous question, although presented in a different way? We have $a_{ij} = 1$ iff $(i, j) \in R$.

@VTand, it is related. Keen observation.

How is $a_2$=6? We have $a_{11}=a_{22}=0$ and two possibilities for each of the two remaining coefficients. These are 4 matrices!

I have already commented this in the other question(which is not just related, it is the same question but phrased different): oeis.org/A001831 You have to tell us what have you tried, if you have read the formulas there and if you get stuck in any step finding the general formula for $n$.

@Phicar I think reading the formulas will not help the OP to understand the problem.

@Phicar, @ miracle, it is helpful.

But I fail to appreciate how this is the number of graded posets with height at most 1?

@FirdousAhmadMala Draw $n$ points and draw a line from $i$ to $j$ if $a_{i,j}=1$. Notice that this is a DAG, so there is a poset associated with the graph being its Hesse diagram. Notice that there are no paths of length greater than $1$, so its maximum chain is $1$ so the height of the poset is at most $1$.

@Phicar, yes. Very revealing.

@Phicar Then probably most clear way to pose the problem is "Find the number of $n\times n$ binary matrices $A$ such that $A^2 = 0$" (due to M. Somos).

@user, I solicit the opinion of Phicar, miracle173, VTand and others.